Thread: Show that the minimized distance from a point to a line is projection onto line

I've a math problem that I like to solve.

Let be a line and let be a point. Show that the orthogonal projection
of onto is the unique point on for which the distance from
to is minimized.


Is it possible to tell me how I can prove this using differential geometry?

I'm not sure how the differential geometry proof will work since I have not studied it yet but what will happen if you take a regular geometric proof and try to translate it into differential geometry lingo?

Originally Posted by Vlasev

I'm not sure how the differential geometry proof will work since I have not studied it yet but what will happen if you take a regular geometric proof and try to translate it into differential geometry lingo?

Yes regular geometric proof is okay. I'll try to translate it into differential geometry.

Why would one even consider "differential geometry" when the Pythagorean theorem gives it easily? In any right triangle, so the hypotenuse is longer than either of the two legs.

Originally Posted by HallsofIvy

Why would one even consider "differential geometry" when the Pythagorean theorem gives it easily? In any right triangle, so the hypotenuse is longer than either of the two legs.

HallofIvy, sorry I didn't understand from your last post. Is it possible to kindly explain how "the hypotenuse is longer than either of the two legs" is related to prove my above stated problem?

Originally Posted by HallsofIvy

Why would one even consider "differential geometry" when the Pythagorean theorem gives it easily? In any right triangle, so the hypotenuse is longer than either of the two legs.

Yes, I got it. Thanks for the answer.